Find the degree of the $P(x)$ Let $P(x)$ be a polynomial and the equality $P(x)=x(1-x)Q(x)$ holds for all real $x$. Suppose that:
$$Q(x)=Q\left(\frac 1{1-x}\right)$$
holds for every $x≠0,1$. Find the degree of the $P(x)$.

My attempts:
We can prove that $Q(x)$ must be a polynomial. Because $$Q(x)=\frac {P(x)}{x(1-x)}$$
Since, $P(x)$ is defined for all $x\in\mathbb R$,this forces $Q(0), Q(1)$ must be defined. Thus, $Q(x)$ is polynomial.
Where $Q(x)$ is undefined for $x=0,1$.
Then, I wrote let $Q(x)=ax+b,a≠0$
$$ax+b=a\left(\frac 1{1-x}\right)+b\implies x=\frac 1{1-x}$$
This shows that, at least $Q(x)≠ax+b$. But, for high degree $Q(x)$ this method gets complicated.
In general, I don't know how to use $$Q(x)=Q\left(\frac 1{1-x}\right)$$

Is it ok starting from:
$$ax+b=a\left(\frac 1{1-x}\right)+b\implies x=\frac 1{1-x}$$
 A: Hint: $\;$ the question is about $\,P(x)\,$, so write the relation in terms of $\,P(x)\,$ alone:
$$
Q(x)=Q\left(\frac 1{1-x}\right)
\;\implies\;
\frac{P(x)}{x(1-x)} = \frac{P\left(\frac{1}{1-x}\right)}{\frac{1}{1-x} \cdot \frac{-x}{1-x}}
\;\implies\;
P(x) = -(1-x)^3 \, P\left(\frac{1}{1-x}\right)
$$

[ EDIT ] $\;$ For verification, $\,\deg P \le 3\,$ in order for the RHS of the last equality to be a polynomial of the same degree with the LHS. The general solution is $\,P(x) = a x^3 - b x^2 - (3 a - b) x + a\,$ with arbitrary $\,a,b \in \mathbb R\,$, as verified here. So, in the end, $\,\deg P = 3\,$ when $\,a \ne 0\,$, or $\,\deg P = 2\,$ when $\,a = 0, b \ne 0\,$, or $\,P(x) \equiv 0\,$ when $\,a=b=0\,$.

[ EDIT #2 ] $\;$ The following addresses the question raised in comments 1, 2 whether the first equality in the problem is supposed to be interpreted as $\,x(1-x) \,\mid\, P(x)\,$, in which case it follows that $\,Q(x)\,$ must be a polynomial, and the second condition trivially implies that it can only be a constant polynomial, meaning $\,P(x)\,$ must be a scalar multiple of $\,x(1-x)\,$.
My reading of the question, on which the answer above was based,  is that $\,Q(x)\,$ is defined by both of the given conditions. The second condition is not defined for $\,x \in \{0, 1\}\,$, which suggests that the domain of $\,Q\,$ should be considered to be $\,\mathbb R \setminus \{0, 1\}\,$, rather than the whole $\,\mathbb R\,$. Under this interpretation, the first condition only applies to the domain of $\,Q\,$, not including $\,\{0,1\}\,$.
Under the alternative interpretation, where the first condition is supposed to hold on the entire $\,\mathbb R\,$, and assuming $\,Q(0), Q(1)\,$ exist and are finite values, it follows that $\,0, 1\,$ must be roots of $\,P(x)\,$, so $\,x(x-1) \,\mid\, P(x)\,$, which is equivalent to $\,a=0\,$ in the posted solution.
A: There are three cases two consider. Either $Q$ is zero, $Q$ is a nonzero polynomial, or $Q$ is a nonzero rational function of polynomials. We consider the first two, as the third case is considered by the other answerer.
Assume that $Q$ is also a polynomial. Note that `$\forall x \in \mathbb R: \left|x\right| > 2 \implies \left| \frac{1}{1-x} \right| \leq 1$ holds. By either using compactness of the interval $I := [-2,2]$ or explicitly using the form of $Q$ being a polynomial admit an absolute bound $B \in \mathbb R^+$ for $Q$ on $I$.
Suppose that $x \in \mathbb R$. If $x \in I$ then $\left|Q(x)\right| \leq B$. Otherwise then $\frac{1}{1 - x} \in I$ and so $\left|Q(x)\right| = \left|Q(\frac{1}{1 - x})\right| \leq B$. Observe now that $B$ bounds $Q$ on $\mathbb R$. $Q$ is a polynomial that is constant on $\mathbb R$ and so is a constant polynomial. Thusly either $Q = 0$ in which case $P$ is also identically zero. Otherwise the degree of $P$ is 2.

In conclusion: either $Q = 0$ hence $P = 0$, or $Q := C \in \mathbb R \setminus \{0\}$, in which case $P = x(1-x) \cdot C$, or $Q$ is a rational function of polynomials. The latter case is covered by the other answer.
